Physics 9 — Friday, December 7, 2018

I Turn in HW11.

I “Practice exam” (due on the last day of class, 12/10) iseffectively a take-home portion of your final exam, intended tohelp you to prepare for the in-class exam (12/17).

I Final exam Mon 12/17 12pm–2pm DRL A6.

I For Wednesday, you read Eric Mazur’s ch 27 (magneticinteractions), which will help us to see how to make electricitydo useful work (turn a motor, ring a doorbell, etc.)

I FYI: positron.hep.upenn.edu/wja/p009/2016/files/exam.pdf

positron.hep.upenn.edu/wja/p009/2016/files/exam_solns.pdf

I Since 14 of you did the (optional/XC) Arduino reading, doesanyone want to work in small groups with Arduinos andbreadboards for a couple of hours during Reading Days?

I https://doodle.com/poll/hwf78raekkknb2eyI Other XC options: read/summarize Muller PTFP chapters:

I 04 – nuclei and radioactivityI 05 – chain reactions, nuclear reactors, and atomic bombsI 11 – quantum physicsI 12 – relativityI 13 – the universeI muller effp ch3.pdf is a slightly more up-to-date (2012)

chapter by Muller on Climate Change

I Or you can read/summarize one or more chapters from anArchitectural Acoustics book by Egan (egan ch1.pdf etc.)

I Or read/summarize 3 Eric Mazur chapters giving a moremathematical look at entropy & thermodynamics.

I Giancoli’s chapter on astronomy (or others if you like)

I Or read+do tutorials on Mathematica for math calculations

I Or code in Processing (java or python): see my Nov 21 slides

Other demos:

I Household light switch with bulb & battery

I Three-way stairway light switch

I swivel magnets & electromagnet & doorbell & speaker

I (in principle) motor & generator

This circuit is more complicated. How many branches? Let’schoose a reference direction for each branch, choose a name forthe current in each branch, and choose a name for all pointsbetween which we might want to measure voltage. (Next page.)

What does junction rule let us write at point c? Point d? Doesthe junction rule at point e tell us anything new?

I count 4 loops. Let’s see what the loop rule tells us. Again, oneequation will be redundant. We’ll just write down the equations,without wasting time to solve them for I1 . . . I5.

I1 = 0.01875A, I2 = 0.0075A, I3 = I4 = 0.00375A, I5 = 0.01125AI1R1 = 1.875V, I2R2 = 0.75V, I3R3 = I4R4 = 0.375V, I5R5 = 1.125V

Wall outlets in a building supply alternating current (AC),whereas a battery supplies direct current (DC). My animation:

https://youtu.be/0wgoPM13kIs

Convention: black=hot, white=neutral, green/bare=ground.

V (t) = Vmax sin(2πft)

Vmax = 170 volts, f = 60 hertz.

Vrms =

√V 2(t) = 120 volts

(Remember that sin2(θ) = 12 so Vrms = Vmax/

√2.)

Wall outlets in a building supply alternating current (AC),whereas a battery supplies direct current (DC). My animation:

https://youtu.be/0wgoPM13kIs

Convention: black=hot, white=neutral, green/bare=ground.

V (t) = Vmax sin(2πft)

Vmax = 170 volts, f = 60 hertz.

Vrms =

√V 2(t) = 120 volts

(Remember that sin2(θ) = 12 so Vrms = Vmax/

√2.)

The usual “junction rule” and “loop rule” work for AC too, butvoltage and current are functions of time now.

In series, each resistor (or lightbulb) gets only a fraction of thetotal emf. So light fixtures arealways in parallel, not in series(except holiday lights, etc.).

E(t)− I (t)R1 − I (t)R2 = 0

Solving for current, we get

I (t) =E(t)

R1 + R2

Voltage across top resistor is

Vba(t) = I (t)R1 =R1

R1 + R2E(t)

V across bottom resistor is

Vcb(t) = I (t)R2 =R2

R1 + R2E(t)

If I plug a 60 watt light bulb into a standard household lightfixture, what is the r.m.s. (root mean square) current that flowsthrough the bulb?

(A) 2.8 amps

(B) 2.0 amps

(C) 1.0 amp

(D) 0.83 amp

(E) 0.5 amp

If I use several “power strips” and several outlets to try to operate100 separate 60-watt lamps simultaneously from the same (120volt r.m.s.) household circuit, how much current (r.m.s.) flowsthrough the fuse or circuit breaker that protects this circuit?

(A) 0.5 amp

(B) 5 amps

(C) 50 amps

What is the r.m.s. current drawn by an 1800-watt hair dryer,operating from a standard 120-volt (r.m.s.) household outlet?

(A) 1.5 amps

(B) 10 amps

(C) 15 amps

(D) 25 amps

A typical U.S. house uses about 1000 kWh per month, i.e. onaverage about 1400 watts, or about 12 amps (at 120 V).

This varies a lot with season and time of day. So a typical U.S.house has 200-amp service coming in from the street, i.e. capableof 24000 watts, in case you have electric stove, electric waterheater, clothes dryer, A/C, hair dryer, etc., all running at once.

Imagine that 500,000 homes in Philadelphia are using 20 amps(r.m.s.) each one evening. That’s 1.2 gigawatts, which is aboutwhat a typical gas/coal/nuclear power plant generates.

At 120 volts, the total current from the power station to the citywould be 107 amps.

In a copper wire of 1 cm radius, the voltage drop after just 2kilometers (R ≈ 0.1 Ω) would be IR ≈ 106 V.

What if the power were instead transmitted at a much highervoltage, like 100 kV ?

Copper wire: 1 cm radius, 2 km length would have R ≈ 0.1 Ω. Wewant to send 109 watts to Philadelphia. What if the power weretransmitted at 100 kV instead of 120 V ?

Need “only” 10,000 amps: (100 kV)(10 kA) = 1 GW.

Voltage drop on 2 km copper wire is then IR = 1 kV, which is only1% of the voltage put out by the power plant. So only 1% of theplant’s power is wasted in heating up the transmission line.

Power delivered is IV — that’s current times voltage. The powerwasted in heating up the cable is I 2Rcable. So to deliver high powerover long distances, we want to keep I as small as possible whilekeeping the product IV large. So very high voltages are used forlong-distance power distribution.

How can you convert from (high current) × (low voltage) into(low current) × (high voltage) and back? A transformer!

Transformer: it only works with AC, not DC!

ES = NSdΦB

dt, EP = NP

dΦB

dt⇒ ES

EP=

NS

NP

Consider an ideal step-down transformer in which the primary coilhas 1000 turns and the secondary coil has 100 turns. The primaryis connected to an AC power source, and the secondary isconnected to a light bulb. If the rms voltage delivered by the ACpower source is 120 volts, what is the rms voltage across the bulb?

(A) 1.2 volts

(B) 12 volts

(C) 120 volts

(D) 1200 volts

Consider an ideal step-down transformer in which the primary coilhas 1000 turns and the secondary coil has 100 turns. The primaryis connected to an AC power source, and the secondary isconnected to a light bulb. If the rms current delivered by the ACpower source is 0.1 amp, what is the rms current through the bulb?

(A) 0.01 amp

(B) 0.1 amp

(C) 1.0 amp

(D) 10 amps

Hint: the power on each side of an ideal transformer is the same.P = IV .

Physics 9 — Friday, December 7, 2018

I Turn in HW11.

I “Practice exam” (due on the last day of class, 12/10) iseffectively a take-home portion of your final exam, intended tohelp you to prepare for the in-class exam (12/17).

I Final exam Mon 12/17 12pm–2pm DRL A6.

I For Wednesday, you read Eric Mazur’s ch 27 (magneticinteractions), which will help us to see how to make electricitydo useful work (turn a motor, ring a doorbell, etc.)

I FYI: positron.hep.upenn.edu/wja/p009/2016/files/exam.pdf

positron.hep.upenn.edu/wja/p009/2016/files/exam_solns.pdf